JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

You throw a ball with a \[\text{\vec{v}}=\left( 3\hat{i}+4\hat{j} \right)\,\,\text{m/s}\] towards a wall, where it hits at height \[{{h}_{1}}\]. Suppose that the launch velocity were, instead, \[\text{\vec{v}}=\left( 5\hat{i}+4\hat{j} \right)\text{m/s}\] and \[{{h}_{2}}\] is height, then

A) \[{{\text{h}}_{\text{1}}}\text{=}{{\text{h}}_{\text{2}}}\]

B) \[{{\text{h}}_{\text{2}}}\text{}{{\text{h}}_{\text{1}}}\]

C) \[{{\text{h}}_{\text{2}}}\text{}{{\text{h}}_{1}}\]

D)        \[{{\text{h}}_{\text{2}}}\ge {{\text{h}}_{\text{1}}}\]

Correct Answer: B

Solution :

[b] \[x=3{{t}_{1}}=5{{t}_{2}}\text{ }\Rightarrow {{t}_{1}}=x/3\text{ and }{{t}_{2}}=x/5\]

Now \[{{h}_{1}}=4{{t}_{1}}-\frac{1}{2}gt_{1}^{2}=\frac{4}{3}x-\frac{g{{x}^{2}}}{18}.\]

and \[{{h}_{2}}=4{{t}_{2}}-\frac{1}{2}gt_{2}^{2}=\frac{4x}{5}-\frac{g{{x}^{2}}}{10}.\]

Clearly, \[{{h}_{2}}<{{h}_{1}}.\]